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CMC Medical CMC-Medical Ludhiana Solved Paper-2014

  • question_answer
    Water from a tap emerges vertically downwards with initial velocity\[4\,m{{s}^{-1}}\]. The cross-sectional area of the tap is A. The flow is steady and pressure is constant throughout the stream of water. The distance h vertically below the tap, where the cross-sectional area of the stream becomes \[\left( \frac{2}{3} \right)\]A, is (g = 10\[m\text{/}{{s}^{2}}\])

    A)  0.5 m                  

    B)  1 m

    C)  1.5 m                  

    D)  2.2 m

    Correct Answer: B

    Solution :

                    The equation of continuity \[{{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}\] \[A\times 4=\frac{2}{3}A\times {{v}_{2}}\]      \[{{v}_{2}}=6\,m{{s}^{-1}}\] From Bernoullis theorem \[p+\rho g{{h}_{1}}+\frac{1}{2}\rho v_{1}^{2}=p+\rho g{{h}_{2}}+\frac{1}{2}\rho v_{2}^{2}\] or            \[g({{h}_{1}}-{{h}_{2}})=\frac{1}{2}(v_{2}^{2}-v_{1}^{2})\] \[g\times h=\frac{1}{2}[{{(6)}^{2}}-{{(4)}^{2}}]\]               \[[\therefore {{h}_{1}}-{{h}_{2}}=h]\] \[10\times h=\frac{1}{2}[36-16]\] \[h=\frac{20}{20}=1\,m\]


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